Problem: The base of a solid $S$ is the region bounded by the curve $y=e^{-x}$, the $x$ -axis, the $y$ -axis, and the line $x=1$. $(0,1)$ $x=1$ ${y=e^{-x}}$ $y$ $x$ Cross-sections perpendicular to the $x$ -axis are equilateral triangles. Determine the exact volume of solid $S$.
Let's graph the base of the solid. The thin orange rectangle depicts a representative cross-section sitting on the base. The length of the green segment is $y$. $y$ $(x,y)$ $(0,1)$ $x=1$ ${y=e^{-x}}$ $y$ $x$ Since each cross-section is perpendicular to the $x$ -axis, the independent variable is $x$. If $A$ denotes the area of each cross-section as a function of $x$, the volume $V$ of solid $S$ is $ V=\int_a^b A(x) \,dx$. To determine the area $A$ as a function of $x$, first express $A$ in terms of $y$. Since the triangular cross-section rests on the rectangle pictured above, the length of the base of the triangle is $y$. Since the triangle is equilateral, the height is $\sqrt3y/2$. $y$ $\dfrac{\sqrt3}2y$ The area $A$ of the triangle is $A=\dfrac12\cdot y\cdot\dfrac{\sqrt3}2y=\dfrac{\sqrt3}4y^2$. What is $A$ as a function of $x$ ? The corner point $(x,y)$ of the rectangle lies on the curve $y=e^{-x}$. We can use this equation to express $A=\dfrac{\sqrt3}4y^2$ in terms of $x$ as $A(x)=\dfrac{\sqrt3}4(e^{-x})^2=\dfrac{\sqrt3}4e^{-2x}$. Can you express the volume $V$ of solid $S$ as a definite integral? Since $x$ goes from $0$ to $1$, the volume formula $ V=\int_a^b A(x) \,dx$ gives us the definite integral $\begin{aligned} V&=\int_0^1 \dfrac{\sqrt3}4e^{-2x}\,dx \\\\ &=\dfrac{\sqrt3}4\int_0^1 e^{-2x}\,dx \end{aligned}$ What is the value of the integral? $\begin{aligned} V&=\dfrac{\sqrt3}4\int_0^1 e^{-2x}\,dx \\\\ &=\dfrac{\sqrt3}4\left[-\dfrac12e^{-2x}\right]_0^1 \\\\ &=\dfrac{\sqrt3}4\left[\dfrac12e^{-2x}\right]^0_1 \\\\ &=\dfrac{\sqrt3}4\cdot\dfrac12\left[e^{-2(0)}-e^{-2(1)}\right] \\\\ &=\dfrac{\sqrt3}8\left(1-e^{-2}\right) \end{aligned}$ Are you wondering what happened to the ${\text{minus sign}}$ in the coefficient of the antiderivative? We got rid of it by switching the order of the limits $0$ and $1$ ! This simple shortcut saves time and removes an opportunity to make a sign error.